FAA2533 wrote: Find all polynomials $P(x)$ for which there exists a sequence $a_1, a_2, a_3, \ldots$ of real numbers such that \[a_m + a_n = P(mn)\]for any positive integer $m$ and $n$. $a_n=P(n)-a_1$ and equation becomes $P(m)+P(n)-2a_1=P(mn)$ This trivially implies $\boxed{P(x)=2a_1}$ constant, which indeed fits (take $a_n=a_1$ $\forall n$)

We have $P(m)=a_m+a_1$. So we can get rid of the sequence and write $P(mn)=P(m)+P(n)-2a_1$. Then, for any fixed $n$, the polynomial $P(nx)-P(x)-P(n)+2a_1$ has infinitely many roots. Thus $P(nx)-P(x)-P(n)+2a_1=0$ for any real $x$. Setting $x=0$ we get $P$ is constant at all infinitely many points ($n=1,2,\dots$), thus $P(x)$ must be a constant polynomial.

sol